Question

Proszę o rozwiązanie zadań w załączniku.

Answer 1

8.

a) W(x) = x²(x² - 2) = x²(x - √2)(x + √2)

b) W(x) = x³(6x² - x + 1)

                      Δ = 1 - 24 = -23 < 0  czyli ta funkcja w nawiasie już nie da się rozłożycz na iloczyn

 

9.

a)

x³ + 5x² - x - 5 = 0

x²(x + 5) - (x + 5) = 0

(x  +5)(x² - 1) = 0

(x + 5)(x - 1)(x + 1) = 0

x + 5 = 0  lub  x - 1 = 0    lub x + 1 = 0

 x = -5      lub    x = 1       lub  x = -1

b)

x² (x - 3)³ ( x + 1) = 0

x² = 0   lub (x - 3)³ = 0  lub  x + 1 = 0

x = 0    lub   x - 3 = 0     lub  x = -1

                     x = 3

c)

x³ - 7x - 6 = 0

x³ - x - 6x - 6 = 0

x(x² - 1) - 6(x + 1) = 0

x(x - 1)(x  +1) - 6(x + 1) = 0

(x + 1)[x(x - 1)  -  6] = 0

x + 1 = 0   lub  x² - x - 6 = 0

 x = -1      lub    Δ = 1 + 24 = 25

                         x1 = (1 - 5)/2 = -4/2 = -2

                         x2 = (1  +5)/2 = 6/2 = 3

odp. x = -1  lub x = -2  lub x = 3

 

d)

-(x² - x) (x² - 4x + 4) = 0

-x(x - 1)(x - 2)² = 0

-x = 0  lub  x - 1 = 0   lub x - 2 = 0

 x = 0    lub x = 1  lub        x = 2

 

11.

y = -1 / x

y = -1 / (x + 1)   + 3 ----- wykres który otrzymamy po przesunięciu

 

D :  x + 1 ≠ 0

      x ≠ -1

D = R \ {-1} --- dziedzina otrzymamnej funkcji

 

12.

a)

3 / (x + 2) + 4 / ( x - 3) = 3(x - 3)/ [(x + 2)(x - 3)] + 4(x + 2) / [(x + 2)(x - 3)] =

(3x - 9 + 4x + 8) / [(x + 2)(x - 3)]  = (7x - 1) / [(x + 2)(x - 3)]

b)

...= (x - 1) / (x + 1) * x²(x + 1) / [(x² - 1)(x²+1)] = x²(x - 1) / [(x - 1)(x + 1)(x² + 1)] = x² / [(x + 1)(x² + 1)

c)

... = (x + 1)² / (x - 4) * [(x - 4)(x + 4)] / (x  +1) = (x + 1)(x  +4) = x² + 5x + 4

 

13

a)

(5x + 4) / (2x - 1) = 3

zał.  2x - 1 ≠ 0

          2x ≠ 1

      D:     x ≠ 1/2

(5x + 4) / (2x - 1) = 3 / 1 

3(2x - 1) = 5x + 4

6x - 3 = 5x + 4

6x - 5x = 4  +3

x = 7 ∈ D

odp. x = 7

 

b)

zał.  2x - 1 ≠ 0   i  x ≠ 0

         x ≠ 1/2

D = R \ {1/2; 0}

 

(x + 1) / (2x - 1) = 2 / x

2(2x  -1) = x(x  +1)

4x - 2 = x² + x

x² + x - 4x + 2 = 0

x² - 3x + 2 = 0

Δ = 9 - 8 = 1

x1 = (3 - 1) / 2 = 2/2 = 1 ∈ D

x2 = (3  +1) / 2 = 4/2 = 2 ∈ D

ODP. X = 1 LUB X = 2

 

14.

a)

zal.  x ≠ 0

2 * x ≤ 0

x ≤ 0   i x ≠ 0

odp. x ∈ (-oo, 0)

 

b)

zał. x ≠ 0

1 / x > - 2

1/x + 2 > 0

1/x + 2x/x > 0

(1 + 2x) / x > 0

(1 + 2x) * x > 0

x(1 + 2x) > 0

x = 0   1 + 2x = 0

                2x = -1

                  x = -1/2

odp. x ∈ (-oo, -1/2) u (0, +oo)

 

Answer 2

Zad.8

a)   W(x) = x^4 - 2x² = x²(x-√2)(x+√2)

b)   W(x) = 6x^5 - x^4 + x^3 = x³(6x²-x+1)

                                                  6x²-x+1 = 0

                                                  Δ = 1-24 = -23 < 0 , nie da się rozłożyć na iloczyn

 

Zad.9

a)

x³+5x²-x-5 = 0

x²(x+5)-(x+5) = 0

(x²-1)(x+5) = (x+5)(x+1)(x-1) = 0

x+5 = 0   => x = -5

lub

x+1 = 0   => x = -1

lub

x-1 = 0   => x = 1

Odp. x = -5   v   x = -1    v    x = 1

b) 

x²(x-3)²*(x+1) = 0

x² = 0   => x = 0

lub

x-3 = 0   => x = 3

lub

x+1 = 0   => x = -1

Odp. x = -1   v   x = 0    v   x = 3

c)

x³-7x-6 = 0

x³-x-6x-6 = 0

x(x²-1)-6(x+1) = 0

x(x+1)(x-1)-6(x1) = 0

(x+1)[x(x-1)-6] = 0

(x+1)(x²-x-6) = 0

x+1 = 0   => x = -1 

lub

x²-x-6 = 0

Δ = 1+24 = 25

√Δ = 5 

x1 = (1-5)/2 = -2

x2 = (1+5)/2 = 3

Odp. x = -2   v   x = -1   v   x = 3

d)

-(x²-x)(x²-4x+4) = 0

-x(x-1)(x-2)² = 0

x = 0

lub

x-1 = 0  => x = 1

lub

x-2 = 0   => x = 2

Odp. x = 0   v   x = 1    v    x = 2

 

Zad.11.

y = -1/x,   u=[-1;3]

y" = -1/(x+1) +3

D: x≠-1

 

Zad,12

a)

3/(x+2) + 4/(x-3) = [3(x-3)+4(x=2)]/[(x+2)(x-3)] = (3x-9+4x+8)/[(x+2(x-3) = 

= (7x-1)/[(x+2)(x-3)]

Z:x ≠ -2

   x ≠ 3

b)

[(x-1)/(x+1)] *(x³+x²)/(x^4 -1) = [(x-1)(x+10] * x²(x=10/[(x²-1)(x²+1)] = 

= x²(x-1)/[(x-1)(x+1)(x²+1)] = x²/[(x=1)(x²+1)]

c)

[(x²+2x+1)/(x-4)] : (x+1)/(x²-16) = [(x+1)²/(x-4)] * [(x-4)(x+4)]/(x+1) =

= (x+1)(x+4) = x²+5x+4

 

Zad.13

a)

(5x+4)/(2x-1) = 3

3(2x-1) = 5x+4

6x-3 = 5x+4

6x-5x = 4+3

x = 7

====

Z: x ≠ 1/2

b)

(x+1)/(2x-1) - 2/x + 0 

(x+1)/(2x-1 = 2/x

x(x+1) = 2(2x-1)

x²+x = 4x-2

x²-3x+2 = 0

Δ = 9-8 = 1

√Δ = 1 

x1 = (3-1)/2 = 1

x2 = (3+1)/2 = 2

 

Z:x ≠ 1/2

   x ≠ 0

 

Zad.14 

a)

2/x ≤ 0 

2*x ≤ 0  /:2

x ≤ 0

 

Z: x ≠ 0 

x < 0

 

b)

1/x > -2

1/x + 2 > 0

(1+2x)/x > 0

x(1+2x) > 0

x+2x² > 0

x(1+2x) > 0

x > 0

lub

1+2x > 0   => x > -1/2

 

Z: x ≠ 0

x ∈ (-∞; -1/2) u (0; +∞)